TITLE: Is this limit indeterminate or $e^2$ or what?
QUESTION [5 upvotes]: What is the answer to this:

$$
\lim_{x\to ∞} \left({2x+3\over 2x-1}\right)^x
$$

My calculator says this is $ e^2 $ but the only answer I can get to is $ 1^\infty $, which is indeterminate.

REPLY [3 votes]: Using the limit definition of the exponential function
$$e^z=\lim_{x\to \infty}\left(1+\frac zx\right)^x$$
we can write
$$\begin{align}
\lim_{x\to \infty}\left(\frac{2x+3}{2x-1}\right)^x&=\lim_{x\to \infty}\left(\frac{1+\frac{3/2}{x}}{1+\frac{-1/2}{x}}\right)^x\\\\
&=\frac{\lim_{x\to \infty}\left(1+\frac {3/2}{x}\right)^x}{\lim_{x\to \infty}\left(1+\frac {-1/2}{x}\right)^x}\\\\\
&=\frac{e^{3/2}}{e^{-1/2}}\\\\
&=e^2
\end{align}$$